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An old antacid commercial claimed that each tablet of their product could neutralize 47 times its mass in stomach acid. The active ingredient in the antacid tablet, NaAl ( OH ) 2 CO 3 , NaAl(OH)2CO3, reacts with the HCl HCl in stomach acid according to the balanced reaction here. NaAl ( OH ) 2 CO 3 + 4 HCl ⟶ NaCl + AlCl 3 + 3 H 2 O + CO 2 NaAl(OH)2CO3+4HCl⟶NaCl+AlCl3+3H2O+CO2 How many moles of HCl HCl can a 1.24 1.24 g antacid tablet neutralize if the tablet contains 0.296 0.296 g of the active ingredient?

Respuesta :

Answer : The number of moles of HCl neutralize is, 0.00824 moles.

Explanation :

The given balanced chemical reaction is:

[tex]NaAl(OH)_2CO_3+4HCl\rightarrow NaCl+AlCl_3+3H_2O+CO_2[/tex]

First we have to calculate the moles of [tex]NaAl(OH)_2CO_3[/tex]

[tex]\text{Moles of }NaAl(OH)_2CO_3=\frac{\text{Mass of }NaAl(OH)_2CO_3}{\text{Molar mass of }NaAl(OH)_2CO_3}[/tex]

Molar mass of [tex]NaAl(OH)_2CO_3[/tex] = 144 g/mol

[tex]\text{Moles of }NaAl(OH)_2CO_3=\frac{0.296g}{144g/mol}=0.00206mol[/tex]

Now we have to calculate the moles of HCl.

From the balanced chemical reaction we conclude that,

As, 1 mole of [tex]NaAl(OH)_2CO_3[/tex] react with 4 moles of HCl

So, 0.00206 mole of [tex]NaAl(OH)_2CO_3[/tex] react with [tex]0.00206\times 4=0.00824[/tex] moles of HCl

Thus, the number of moles of HCl neutralize is, 0.00824 moles.

pstnonsonjoku

0.0084 moles of HCl is required to neutralize the antacid tablet.

The equation of the reaction is;  

NaAl(OH)2CO3 + 4HCl ⟶NaCl + AlCl3 + 3H2O + CO2

Molar mass of antacid = 144 g/mol

Number of moles of antacid =  0.296 g/144 g/mol = 0.0021 moles

From the reaction equation, we can see that the reaction is 1:4

1 mole of antacid reacts with 4 moles of HCl

0.0021 moles of antacid reacts with 0.0021 moles × 4 moles/1 mole

= 0.0084 moles

Hence, 0.0084 moles of HCl is required to neutralize the antacid tablet.

Learn more: https://brainly.com/question/2510654

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